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I've been beating around the bush for a long time, ever since I've been in StackExchange. Now, I really must come to the point. I have a link here that you people can see (go to about the end of the eighth page)

Given a value for $p$, how will you generate the sequence, moreover, what will the $n$th element in the set be?

Edit Perhaps I should've written this before, never mind... I meant to say, how do I go about generating a unique-sum set with $p$ elements, as the few examples on page 9 (i.e. following part 2.3) show, and when done so, how would I know what the $n$th element in that set would be, considering I won't be looking over the entire thing once I'm done generating the set?

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Isn't there some way you can explain your question without asking your audience to first read a 16-page paper? – MJD Apr 4 '12 at 17:00
Aside from the word set in the title, what is the relation to set theory? Can it be understood without reading the said 16 pages long paper? – Asaf Karagila Apr 4 '12 at 17:04
The relevant content of the paper is all in section 2.3: a "unique-sum set" of positive integers is a set $\{s_1...s_n\}$ for which the only solution to $\sum c_is_i = \sum s_i$ has $c_i = 1$ for each $i$. But I cannot make out what question is being asked. – MJD Apr 4 '12 at 17:07
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The first half of the post mentions beating around the bush and getting to the point. The second half is evasive and unclear. Ironic, no? – Théophile Apr 4 '12 at 17:15
@Mark Dominus: Probably not. It's hard to put in words, and I'm terrible in TeX. Sorry for that. Moreover, you get to see more references and sources within the entire paper. – Mach9 Apr 4 '12 at 17:37
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In the example on page 9, the set $U_p$ contains the $p$ numbers $2^p - 2^{p-m}$ for each $m$ in $1\ldots p$. For example, $U_5 = \{ 32-1, 32-2, 32-4, 32-8, 32-16 \} = \{ 16, 24, 28, 30, 31 \}$. Proposition 2 claims that the $U_p$ sets have the unique-sum property. Is this what you were looking for?

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That effectively answers it-thanks, Mr. Mark Dominus. The paper wasn't all that clear to me. – Mach9 Apr 4 '12 at 18:49

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